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GR9277 #28
Problem
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\prob{28}
A system is known to be in the normalized state described by the wave function
,
where the are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m=3 is

1. 0
2. 1/15
3. 1/6
4. 1/3
5. 13/15

Quantum Mechanics}Probability

One doesn't actually need to know much (if anything) about spherical harmonics to solve this problem. One needs only the relation . Since the problem asks for states where , and it gives the form of spherical harmonics employed as , one can eliminate the third term after the dot-product.

So, the given wave function gets dot-product'ed like , as in choice (E).  Alternate Solutions
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mpdude8
2012-04-19 19:55:26
One can also deduce that the particle is more likely than not to have m = 5 by looking at the normalized wave function. E is the only possibility that fits. tera
2006-08-21 04:37:41
No dirac symbols not anything. The Possibility is c1^2 + c2^2 as ina a simple mathematical problem in our case 25/30+1/30=26/30
 jmason862009-09-09 22:12:22 agreed, the probability of finding a system in a given state is always just the sum of the squares of the coefficients in the wavefunction.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$